CK-12-Calculus

http://www.ck12.org Chapter 7. Integration Techniques

Evaluate∫xsinxdx.
Solution:
We use the formula∫udv=uv−∫vdu.
Choose

u=x

and

dv=sinxdx.

To complete the formula, we take the differential ofuand the simplest antiderivative ofdv=sinxdx.

du=dx

v=−cosx.

The formula becomes

∫

xsinxdx=−xcosx−

∫

(−cosx)dx

=−xcosx+

∫

cosxdx

=−xcosx+sinx+C.

A Guide to Integration by Parts
Which choices ofuanddvlead to a successful evaluation of the original integral? In general, chooseuto be
something that simplifies when differentiated, anddvto be something that remains manageable when integrated.
Looking at the example that we have just done, we choseu=xanddv=sinxdx.That led to a successful evaluation
of our integral. However, let’s assume that we made the following choice,

u=sinx

dv=xdx.

Then

du=cosxdx

v=x^2 / 2.

Substituting back into the formula to integrate, we get

∫

udv=uv−

∫

vdu

=sinxx

2

2 −

∫ x 2

2 cosxdx